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Math Help - Potential difference

  1. #1
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    Potential difference

    If we're given V=2x^2+2y^2 and we're told to find the field at a point x=3, y=3, would this method be correct?

    Ex=d(2x^2)/dx=4x=12

    Ey=d(-2y^2)/dy=-4y=-12

    Therefore, E={12i - 12j}

    Not entirely sure, this is the first of its kind.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by quantoembryo View Post
    If we're given V=2x^2+2y^2 and we're told to find the field at a point x=3, y=3, would this method be correct?

    Ex=d(2x^2)/dx=4x=12

    Ey=d(-2y^2)/dy=-4y=-12

    Therefore, E={12i - 12j}

    Not entirely sure, this is the first of its kind.
    Mostly correct. E = - \nabla V so you are off by a minus sign.

    -Dan
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  3. #3
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    So then it would be E={-12i+12j}? That actually makes a lot more sense.
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  4. #4
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    Do you have a typo in your potential?

    You wrote in your post that: V=2x^2+2y^2\,.

    You keep responding as if V=2x^2-2y^2\,.
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  5. #5
    Forum Admin topsquark's Avatar
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    Good catch. I missed that.

    -Dan
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  6. #6
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    Yes, that was a typo. Sorry! Thanks for the help! One quick question pertaining to that relationship you posted. What is that greek symbol and what does it mean? Thanks
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  7. #7
    A Plied Mathematician
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    If you are referring to the \nabla in the topsquark's expression E = - \nabla V, then it's not a Greek letter, actually. It's called the "del" operator, or also the "nabla". In this context, it indicates that you are to take the gradient of the scalar function V.
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  8. #8
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    Alright, I have yet to encounter that sign so it's probably best if I say E=-dv/dx correct?
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  9. #9
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    Alright, I have yet to encounter that sign so it's probably best if I say E=-dv/dx correct?
    Not true for your problem, actually. If you were working in only one dimension, it would be true. For your case, you should have the vector equation

    \mathbf{E}=-\left\langle\dfrac{\partial V}{\partial x},\dfrac{\partial V}{\partial y},\dfrac{\partial V}{\partial z}\right\rangle.

    Or, for two dimensions, you'd have

    \mathbf{E}=-\left\langle\dfrac{\partial V}{\partial x},\dfrac{\partial V}{\partial y}\right\rangle.
    Last edited by Ackbeet; February 10th 2011 at 06:23 AM. Reason: Two Dimensions.
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  10. #10
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    Alright, let me rephrase. Ex=-dV/dx, and Ey=-dV/dy
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  11. #11
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    As long as your "d's" mean "partial derivative" and not "ordinary derivative", you're ok there.
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